Derivative f’ of function f(x)=arctan x is: f’(x) = 1 / (1 + x²) for all x real. To show this result, we use derivative of the inverse function tan x.

Derivative of arctan x

Derivative $f’$ of function $f(x)=\arctan{x}$ is: $$\forall x \in \mathbb{R} ,\quad f'(x) = \dfrac{1}{1+x^2}$$

Proof

Remember that function $\arctan$ is the inverse function of $\tan$ :

$\left(f^{-1} \circ f\right)=\left(\tan \circ \arctan\right)(x)=\tan(\arctan(x))=x$

Using result of derivative of inverse functions+ we have:

$(g^{-1})^{\prime}(x)=\frac{1}{g^{\prime}(g^{-1}(x))}$

Taking : $g^{-1}=f=\arctan$ and $g=f^{-1}=\tan$, we have: $$f^{\prime}(x)=\frac{1}{\tan^{\prime}(f(x))}$$

We have:

Derivative of tangent function $g(x)=\tan x$ is: $$\forall x \neq \dfrac{\pi}{2}+k\pi, k \in \mathbb{Z}, \quad g'(x) = 1+\tan ^{2} x$$

So, we have:

\begin{aligned} f^{\prime}(x)&=\frac{1}{\tan^{\prime}(f(x))}\\ &=\frac{1}{1+\tan^2(f(x))}\\ &=\frac{1}{1+\tan^2(\arctan x)}\\ \end{aligned}

Now by defiinition: $$\left(f^{-1} \circ f\right)=\left(\tan \circ \arctan\right)(x)=\tan(\arctan(x))=x$$

We conclude that: $$\forall x \in \mathbb{R} ,\quad f'(x) = \dfrac{1}{1+x^2}$$